Higher: scale and reflect sine graphs in degrees
Separate output scaling from input scaling, recover signed factors from labelled landmarks, and draw the valid transformed graphs.
Section 3.3 · Graphs (teaching linked; review pending)
Exam scope and assessment notes
Additional Higher only. Higher includes the shared Foundation content. The Graphs route has nine shared lessons, two Higher line-equation lessons, four Higher graph-family lessons, then these nine transformation lessons. H:3.3B requires all eight core transformation lessons: all four forms f(x)+a, f(x+a), af(x), f(ax) across linear, quadratic, degree sine and degree cosine graphs, with worked construction and every practice stage. H:3.3C requires equation recovery and analysis in all eight core lessons plus the ninth domain/equivalence/order synthesis lesson. This nine-lesson split is StudyForA’s teaching design, not a syllabus-prescribed lesson count. Map landmarks, write equations and draw the original and both target graphs on labelled axes; an automatic trace alone does not replace these constructions. Finite interval endpoints in these exercises are included and infinite bounds are not. Retain input preimages and exclusions; zero factors, even or periodic formulas do not restore invalid arguments or establish a unique history. The bounded interactive examples support exploration, not a general expression or composition parser. All nine Foundation and seven additional Higher Graphs source rows have teaching and staged-practice links. These links do not establish exhaustive subskill approval, mastery or examination readiness; specialist review and formal assessment validation remain pending. Written work is not automatically marked; explained answers support self-checking, not exam marks or proof of mastery. Specialist review and formal assessment validation remain unestablished. Pearson chapter-map and uncorrected-proof navigation remain separate evidence; final full-book and Teaching Hub alignment has not been verified.
Before you begin
- Evaluate the original sine rule and plot exact coordinate pairs.
- Read function notation and solve simple equations with signed fractions.
- Use interval domains and ranges and standard degree sine/cosine values.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For y=3sin(2x), which amplitude/period pair is correct?
What you will learn
- Apply af(x) and f(ax), with nonzero a to sine graphs using exact point mappings.
- Recover unknown parameters and transformed equations from known forms and labelled correspondences.
- Draw the original and both targets, retaining exact domains and ranges and distinguishing amplitude from the full-rule period measured in degrees.
Original: f(x) = sin(x°). Target: y = af(x), with a = 2. All angles and horizontal shifts use degrees.
The domain choice changes the original exercise deliberately. Every finite interval endpoint here is included; an infinite bound is never included.
Applied equation: y = 2 × sin(x°). Original domain: All real numbers; target domain: All real numbers. Original range: [-1, 1]; target range: [-2, 2].
Point rule: (u,v) → (u, (2)v).
The extending target formula has amplitude 2 and least positive period 360°. The all-real target has that period.
Blue dashed: original. Red solid: target. Filled points mark supplied landmarks or included endpoints; grey dotted joins show point correspondences, not additional curves. If traces coincide, the red trace can cover the blue. The window is x from -360 to 360, y from -5 to 5; it does not change either mathematical domain.
At target input x=0, use original argument 0. Exact target output = 0. The selected target point is ringed in white.
Show transformation values and mapped landmarks
| Target x | Original argument | Target output |
|---|---|---|
| -180 | -180 | 0 |
| -90 | -90 | -2 |
| 0 | 0 | 0 |
| 30 | 30 | 1 |
| 60 | 60 | (2) × (√3/2) |
| 90 | 90 | 2 |
| 180 | 180 | 0 |
| Original (u,v) | Image (x,y) |
|---|---|
| (30, 1/2) | (30, 1) |
| (90, 1) | (90, 2) |
Exact fractions and standard-angle expressions may remain unsimplified. Other sine/cosine readings are marked approximate. The curve is a numerical illustration; your own drawn graph readings are approximate. Use the point correspondences, equation and domain together to check your construction.
Separate sine amplitude from a signed output factor
For y=af(x), multiply every sine height by a and keep each degree input. A negative factor reflects in the x-axis; amplitude is nonnegative, so the amplitude of aC sin x is |aC|. The full-rule period remains 360° for these nonzero output factors, and the midline stays y=0.
Map zeros and peaks as well as the supplied half-height points. An old zero cannot recover an output factor because every factor sends it to zero. Use a nonzero labelled height, calculate a=v_new/v_old, then check the second correspondence. Draw a smooth wave through the resulting exact landmarks.
Change the sine input and its period
For y=f(ax), solve ax=u, giving (u,v)→(u/a,v). The amplitude stays |C|, while the full-rule period becomes 360°/|a|. A negative a reflects in the y-axis as well as scaling horizontal coordinates; a factor of magnitude below 1 makes the horizontal spacing larger.
Sine is odd: sin(−x)=−sin x on the full real degree domain. Thus an input reflection and an output factor −1 can describe the same full trace, though their point mappings differ. Do not confuse this identity with a general statement that input and output factors are interchangeable: their effects on period and amplitude usually differ.
Keep degree scales and restricted preimages
Draw the original and both targets on degree axes with space for their mapped peaks and zeros. Use amplitude and period to add suitable landmarks rather than joining a few points by straight segments. The equations fix those coordinates exactly; estimates read elsewhere from the drawing are approximate.
With a restricted original interval, require ax to lie in that interval and reverse inequalities when dividing by a negative number. The range is unchanged under a nonzero input scale, but the allowed horizontal interval changes. A period such as 720° describes the full extended rule, not a claim that a finite restricted piece repeats globally.
Sources for this lesson
Pause and explain
Which full real degree identity is correct?
Worked example
Start from f(x)=sin x, with domain all real x (in degrees) and range [-1, 1]. Original landmarks: (90, 1), (30, 1/2). A uses y=af(x) with a=2: an output scale of magnitude 2. B uses y=f(ax) with a=2: a horizontal coordinate scale of 1/2. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point. Use degree axes, and state amplitude, midline and the period of each underlying full rule.
Show the worked solution
- A: an output scale of magnitude 2. Use (u,v) → (u,av) with a=2. The labelled points become (90, 1) → (90, 2); (30, 1/2) → (30, 1). B: a horizontal coordinate scale of 1/2. Use (u,v) → (u/a,v), with a≠0 with a=2. The labelled points become (90, 1) → (45, 1); (30, 1/2) → (15, 1/2).
- A: y=2sin(x); all angles in degrees. Domain all real x (in degrees); range [-2, 2]. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. Amplitude 2, midline y=0, underlying full-rule period 360°. B: y=sin(2x); all angles in degrees. Domain all real x (in degrees); range [-1, 1]. Require (2)x in the original domain; divide its endpoints by the parameter and reorder them when it is negative. Amplitude 1, midline y=0, underlying full-rule period 180°.
- Check the extra original point (-90, -1): A gives (-90, -2) and B gives (-45, -1). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate. Include consecutive repeated full-rule landmarks when choosing the view, or mark how the restricted endpoints prevent a complete repeat.
Answer A: y=2sin(x); all angles in degrees; D=all real x (in degrees), R=[-2, 2]. Amplitude 2, underlying period 360°. B: y=sin(2x); all angles in degrees; D=all real x (in degrees), R=[-1, 1]. Amplitude 1, underlying period 180°.
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Start from f(x)=sin x, with domain all real x (in degrees) and range [-1, 1]. Original landmarks: (30, 1/2), (90, 1). A uses y=af(x) with a=-3: an output scale of magnitude 3 followed by reflection in the x-axis. B uses y=f(ax) with a=-1: a horizontal coordinate scale of 1 together with reflection in the y-axis. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point. Use degree axes, and state amplitude, midline and the period of each underlying full rule. For B, sin(−x)=−sin x on the full real degree domain. This input reflection has the same trace as an output factor −1, while A has output factor −3 and a different amplitude.
Give me a hint
Track one coordinate at a time. For an input change set the new argument equal to the old input, then solve for the new horizontal coordinate. Apply the same rule to the domain endpoints.
Compare my reasoning
- A: an output scale of magnitude 3 followed by reflection in the x-axis. Use (u,v) → (u,av) with a=-3. The labelled points become (30, 1/2) → (30, -3/2); (90, 1) → (90, -3). B: a horizontal coordinate scale of 1 together with reflection in the y-axis. Use (u,v) → (u/a,v), with a≠0 with a=-1. The labelled points become (30, 1/2) → (-30, 1/2); (90, 1) → (-90, 1).
- A: y=-3sin(x); all angles in degrees. Domain all real x (in degrees); range [-3, 3]. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. Amplitude 3, midline y=0, underlying full-rule period 360°. B: y=sin(−x); all angles in degrees. Domain all real x (in degrees); range [-1, 1]. Require (-1)x in the original domain; divide its endpoints by the parameter and reorder them when it is negative. Amplitude 1, midline y=0, underlying full-rule period 360°.
- Check the extra original point (-90, -1): A gives (-90, 3) and B gives (90, -1). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate. For B, sin(−x)=−sin x on the full real degree domain. This input reflection has the same trace as an output factor −1, while A has output factor −3 and a different amplitude. Include consecutive repeated full-rule landmarks when choosing the view, or mark how the restricted endpoints prevent a complete repeat.
A: y=-3sin(x); all angles in degrees; D=all real x (in degrees), R=[-3, 3]. Amplitude 3, underlying period 360°. B: y=sin(−x); all angles in degrees; D=all real x (in degrees), R=[-1, 1]. Amplitude 1, underlying period 360°.
Look for these in your work
- Draw and label the original plus both targets, with all supplied and one extra point mapped correctly.
- Give both exact equations and their actual domains and ranges, preserving restricted endpoints.
- Use degrees and separate amplitude from period; qualify periods of restricted portions.
Start from f(x)=2sin x, with domain all real x (in degrees) and range [-2, 2]. Original landmarks: (90, 2), (30, 1). A is known to have form y=af(x); its labelled correspondences are (90, 2) → (90, 1); (30, 1) → (30, 1/2). B is known to have form y=f(ax); its labelled correspondences are (90, 2) → (180, 2); (30, 1) → (60, 1). Recover each unknown a from those correspondences and explain why the given form matters. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point. Use degree axes, and state amplitude, midline and the period of each underlying full rule.
Give me a hint
Use the known form to decide which coordinate is preserved. Recover an output shift by a difference, an input shift from u=x+a, an output scale from y=av, or an input scale from u=ax; choose a nonzero coordinate for division.
Compare my reasoning
- A: Using (90, 2) → (90, 1), a=y/v=(1)/(2)=1/2. Use (u,v) → (u,av) with a=1/2. The labelled points become (90, 2) → (90, 1); (30, 1) → (30, 1/2). B: Using (90, 2) → (180, 2), u=ax, so a=u/x=(90)/(180)=1/2. Use (u,v) → (u/a,v), with a≠0 with a=1/2. The labelled points become (90, 2) → (180, 2); (30, 1) → (60, 1).
- A: y=sin(x); all angles in degrees. Domain all real x (in degrees); range [-1, 1]. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. Amplitude 1, midline y=0, underlying full-rule period 360°. B: y=2sin((1/2)x); all angles in degrees. Domain all real x (in degrees); range [-2, 2]. Require (1/2)x in the original domain; divide its endpoints by the parameter and reorder them when it is negative. Amplitude 2, midline y=0, underlying full-rule period 720°.
- Check the extra original point (-90, -2): A gives (-90, -1) and B gives (-180, -2). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate. Include consecutive repeated full-rule landmarks when choosing the view, or mark how the restricted endpoints prevent a complete repeat.
A: y=sin(x); all angles in degrees; D=all real x (in degrees), R=[-1, 1]. Amplitude 1, underlying period 360°. B: y=2sin((1/2)x); all angles in degrees; D=all real x (in degrees), R=[-2, 2]. Amplitude 2, underlying period 720°.
Look for these in your work
- Draw and label the original plus both targets, with all supplied and one extra point mapped correctly.
- Give both exact equations and their actual domains and ranges, preserving restricted endpoints.
- Recover each unknown parameter from labelled correspondences and the stated form.
Start from f(x)=sin x, with domain [-90°, 90°] and range [-1, 1]. Original landmarks: (30, 1/2), (90, 1). A uses y=af(x) with a=-2: an output scale of magnitude 2 followed by reflection in the x-axis. B uses y=f(ax) with a=-1/2: a horizontal coordinate scale of 2 together with reflection in the y-axis. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point. Use degree axes, and state amplitude, midline and the period of each underlying full rule. Include the finite endpoints. Explain why extending a drawing or rewriting a formula does not restore an excluded original input.
Give me a hint
Track one coordinate at a time. For an input change set the new argument equal to the old input, then solve for the new horizontal coordinate. Apply the same rule to the domain endpoints.
Compare my reasoning
- A: an output scale of magnitude 2 followed by reflection in the x-axis. Use (u,v) → (u,av) with a=-2. The labelled points become (30, 1/2) → (30, -1); (90, 1) → (90, -2). B: a horizontal coordinate scale of 2 together with reflection in the y-axis. Use (u,v) → (u/a,v), with a≠0 with a=-1/2. The labelled points become (30, 1/2) → (-60, 1/2); (90, 1) → (-180, 1).
- A: y=-2sin(x); all angles in degrees. Domain [-90°, 90°]; range [-2, 2]. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. Amplitude 2, midline y=0, underlying full-rule period 360°. This finite restricted graph has no global repetition domain; the stated period belongs to its full extension. B: y=sin((-1/2)x); all angles in degrees. Domain [-180°, 180°]; range [-1, 1]. Require (-1/2)x in the original domain; divide its endpoints by the parameter and reorder them when it is negative. Amplitude 1, midline y=0, underlying full-rule period 720°. This finite restricted graph has no global repetition domain; the stated period belongs to its full extension.
- Check the extra original point (-90, -1): A gives (-90, 2) and B gives (180, -1). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate. The endpoint images for A are (-90, -1) → (-90, 2); (90, 1) → (90, -2). The endpoint images for B are (-90, -1) → (180, -1); (90, 1) → (-180, 1). Include consecutive repeated full-rule landmarks when choosing the view, or mark how the restricted endpoints prevent a complete repeat.
A: y=-2sin(x); all angles in degrees; D=[-90°, 90°], R=[-2, 2]. Amplitude 2, underlying period 360°. B: y=sin((-1/2)x); all angles in degrees; D=[-180°, 180°], R=[-1, 1]. Amplitude 1, underlying period 720°.
Look for these in your work
- Draw and label the original plus both targets, with all supplied and one extra point mapped correctly.
- Give both exact equations and their actual domains and ranges, preserving restricted endpoints.
- Use degrees and separate amplitude from period; qualify periods of restricted portions.
Common mistakes
- Multiplying horizontal coordinates by a for f(ax) instead of dividing by nonzero a.
- Dropping a constant under output scaling, or failing to reverse finite bounds for negative input factors.
- Treating an unchanged even trace as an unchanged point mapping or treating a signed coefficient as a negative amplitude or period.
What can you explain now?
For y=−2sin(x/3), state amplitude and the least positive full-rule period in degrees.
Compare with the explanation
Amplitude 2; period 1080°.
Take the absolute output coefficient and divide 360° by the magnitude 1/3 of the input factor.
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Make it stick
Explain the input and output mappings, recover a parameter from a labelled point, and state how a restricted domain survives the change. State amplitude and period separately, using degrees.
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